Bayesian Batting Average Prior

Question

I wanted to ask a question inspired by an excellent answer to the query about the intuition for the beta distribution. I wanted to get a better understanding of the derivation for the prior distribution for the batting average. It looks like David is backing out the parameters from the mean and the range.

Under the assumption that the mean is $0.27$ and the standard deviation is $0.18$, can you back out $\alpha$ and $\beta$ by solving these two equations: \begin{equation} \frac{\alpha}{\alpha+\beta}=0.27 \\ \frac{\alpha\cdot\beta}{(\alpha+\beta)^2\cdot(\alpha+\beta+1)}=0.18^2 \end{equation}

Answer 1

Notice that:

\begin{equation} \frac{\alpha\cdot\beta}{(\alpha+\beta)^2}=(\frac{\alpha}{\alpha+\beta})\cdot(1-\frac{\alpha}{\alpha+\beta}) \end{equation}

This means the variance can therefore be expressed in terms of the mean as

\begin{equation} \sigma^2=\frac{\mu*(1-\mu)}{\alpha+\beta+1} \\ \end{equation}

If you want a mean of $.27$ and a standard deviation of $.18$ (variance $.0324$), just calculate:

\begin{equation} \alpha+\beta=\frac{\mu(1-\mu)}{\sigma^2}-1=\frac{.27\cdot(1-.27)}{.0324}-1=5.083333 \\ \end{equation}

Now that you know the total, $\alpha$ and $\beta$ are easy:

\begin{equation} \alpha=\mu(\alpha+\beta)=.27 \cdot 5.083333=1.372499 \\ \beta=(1-\mu)(\alpha+\beta)=(1-.27) \cdot 5.083333=3.710831 \end{equation}

You can check this answer in R:

> mean(rbeta(10000000, 1.372499, 3.710831))
[1] 0.2700334
> var(rbeta(10000000, 1.372499, 3.710831))
[1] 0.03241907